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Question Number 193426 by 073 last updated on 13/Jun/23

$$\int \sqrt[3]{{\mathrm{x}}^3+1}\mathrm{dx}=?$$$$\mathrm{solution}?$$

Answered by witcher3 last updated on 15/Jun/23

$${\mathrm{x}}^3=\mathrm{t}$$$$\int {({\mathrm{x}}^3+1)}^{\frac{1}{3}}\mathrm{dx}=\mathrm{f}\left(\mathrm{x}\right)$$$$\frac{1}{3}\int {(\mathrm{t}+1)}^{\frac{1}{3}}{\mathrm{t}}^{-\frac{2}{3}}\mathrm{dt}$$$${(\mathrm{t}+1)}^{\frac{1}{3}}=1+\sum _{\mathrm{k}⩾1}\frac{\underset{\mathrm{j}=1}{\prod ^{\mathrm{k}}}(\frac{4}{3}-\mathrm{j})}{\left(\mathrm{k}\right)!}{\mathrm{t}}^{\mathrm{k}}$$$$\frac{1}{3}\int {\mathrm{t}}^{-\frac{2}{3}}+\sum _{\mathrm{k}⩾1}\frac{{(-1)}^{\mathrm{k}}\mathrm{\Gamma }(\mathrm{k}-\frac{1}{3})}{\mathrm{\Gamma }(\mathrm{k}+1)\mathrm{\Gamma }(-\frac{1}{3})}{\mathrm{t}}^{\mathrm{k}-\frac{2}{3}}\mathrm{dt}$$$$={\mathrm{t}}^{\frac{1}{3}}+\frac{1}{3}\sum _{\mathrm{k}⩾1}\frac{\mathrm{\Gamma }(\mathrm{k}-\frac{1}{3}){(-1)}^{\mathrm{k}}}{\mathrm{\Gamma }(-\frac{1}{3})(\mathrm{k}+\frac{1}{3})}\frac{{\mathrm{t}}^{\mathrm{k}+\frac{1}{3}}}{\mathrm{k}!}+\mathrm{c}$$$${\mathrm{t}}^{\frac{1}{3}}(1+\frac{1}{3}\sum _{\mathrm{k}⩾1}\frac{\mathrm{\Gamma }(\mathrm{k}-\frac{1}{3})}{\mathrm{\Gamma }(-\frac{1}{3})}.\frac{\mathrm{\Gamma }(\mathrm{k}+\frac{1}{3})}{\mathrm{\Gamma }(\mathrm{k}+\frac{4}{3})}.\frac{{(-\mathrm{t})}^{\mathrm{k}}}{\mathrm{k}!})+\mathrm{c}$$$$\frac{1}{3}=\frac{\mathrm{\Gamma }\left(\frac{4}{3}\right)}{\mathrm{\Gamma }\left(\frac{1}{3}\right)}$$$$\iff {\mathrm{t}}^{\frac{1}{3}}(1+\sum _{\mathrm{k}⩾1}\frac{\mathrm{\Gamma }(\mathrm{k}-\frac{1}{3})}{\mathrm{\Gamma }(-\frac{1}{3})}.\frac{\frac{\mathrm{\Gamma }(\mathrm{k}+\frac{1}{3})}{\mathrm{\Gamma }\left(\frac{1}{3}\right)}}{\frac{\mathrm{\Gamma }(\mathrm{k}+\frac{4}{3})}{\mathrm{\Gamma }\left(\frac{4}{3}\right)}}.\frac{{(-\mathrm{t})}^{\mathrm{k}}}{\mathrm{k}!})+\mathrm{c}$$$${\left(\mathrm{a}\right)}_{\mathrm{k}}=\frac{\mathrm{\Gamma }(\mathrm{a}+\mathrm{k})}{\mathrm{\Gamma }\left(\mathrm{a}\right)}$$$${\mathrm{t}}^{\frac{1}{3}}(1+\sum _{\mathrm{k}⩾1}\frac{{(-\frac{1}{3})}_{\mathrm{k}}{\left(\frac{1}{3}\right)}_{\mathrm{k}}}{{\left(\frac{4}{3}\right)}_{\mathrm{k}}}.\frac{{(-\mathrm{t})}^{\mathrm{k}}}{\mathrm{k}!})+\mathrm{c}$$$${\mathrm{t}}^{\frac{1}{3}}{}_2{\mathrm{F}}_1(-\frac{1}{3},\frac{1}{3};\frac{4}{3},-\mathrm{t})+\mathrm{c}$$$$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}_2{\mathrm{F}}_1(-\frac{1}{3},\frac{1}{3};\frac{4}{3},-{\mathrm{x}}^3)+\mathrm{c},\mathrm{c}\in \mathbb{R}$$$$$$$$$$$$$$

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